英文:
Get a list of keys that match each item of another given list...?
问题
result_list = [key for key, values in my_dict.items() if any(item in values for item in search_list)]
英文:
my_dict = {'NONE': ['N', 'NE'],'VERY SLIGHT': ['VSLT', 'VT'],'SLIGHT': ['SLT', 'ST'],'FAINT': ['F', 'FT'],'MEDIUM': ['M', 'MM'],'STRONG': ['S', 'SG'],'VERY STRONG': ['VS', 'VG']}search_list = ['N', 'SLT', 'MM']result_list = []
I want to search each item of search_list in values of my_dict and return the matching keys as list.
I need... result_list = ['NONE', 'SLIGHT', 'MEDIUM']
I got it by...
for i in search_list:for k, v in my_dict.items():if i in v:result_list.append(k)
how to get it in more simpler and quickest way to do it?
答案1
得分: 1
将文本到代码的字典翻转为代码到文本的字典（每个程序运行一次），然后只需进行简单快速的字典查找。
英文:
"Flip" your text-to-codes dict to a code-to-text dict (once per your program), and then it's just a simple, fast dict lookup away.
text_to_codes = {"NONE": ["N", "NE"],"VERY SLIGHT": ["VSLT", "VT"],"SLIGHT": ["SLT", "ST"],"FAINT": ["F", "FT"],"MEDIUM": ["M", "MM"],"STRONG": ["S", "SG"],"VERY STRONG": ["VS", "VG"],}code_to_text = {}for text, codes in text_to_codes.items():code_to_text.update({code: text for code in codes})# `code_to_text` will end up looking like# {'N': 'NONE', 'NE': 'NONE', 'VSLT': 'VERY SLIGHT', 'VT': 'VERY SLIGHT', 'SLT': 'SLIGHT', ...# at this point.search_list = ["N", "SLT", "MM"]result_list = [code_to_text[code] for code in search_list]print(result_list) # ['NONE', 'SLIGHT', 'MEDIUM']
答案2
得分: 0
嵌套推导是一种选择。
[k for k, v in my_dict.items() for x in search_list if x in v]
英文:
Nested comprehension is one option.
[k for k, v in my_dict.items() for x in search_list if x in v]
答案3
得分: 0
Sure, here's the translated code:
search_list = ['N', 'SLT', 'MM']find_list = []for k, v in my_dict.items():find_list.extend(map(lambda x: (x, k), v))dict_result = {tup[0]: tup[1:] for tup in find_list}for i in search_list:print(dict_result[i])
Please note that I've translated the code as requested, but I recommend using a more meaningful variable name than "dict" to avoid conflicts with the built-in Python dict type.
英文:
search_list = ['N', 'SLT', 'MM']find_list=[]for k,v in my_dict.items():find_list.extend(map(lambda x: (x,k), v))dict = {tup[0]: tup[1:] for tup in find_list}for i in search_list:print(dict[i])
答案4
得分: -1
You could use list comprehension:
search_list = ["N", "SLT", "MM"]result_list = [k for k, v in my_dict.items() for x in search_list if x in v]
This adds the appropriate keys to the result list.
(Add a key only if one of its items is the abbreviation you want.)
On the other hand, you could flip your dictionary, then just look it up:
code_to_text = {}for k, v in text_to_codes.items():code_to_text.update({code: k for code in v})search_list = ["N", "SLT", "MM"]result_list = [code_to_text[code] for code in search_list]
Both of these give the same value of result_list:
['NONE', 'SLIGHT', 'MEDIUM']
英文:
You could use list comprehension:
search_list = ["N", "SLT", "MM"]result_list = [k for k, v in my_dict.items() for x in search_list if x in v]
This adds the appropriate keys to the result list.
(Add a key only if one of its items is the abbreviation you want.)
On the other hand, you could flip your dictionary, then just look it up:
code_to_text = {}for k, v in text_to_codes.items():code_to_text.update({code: k for code in v})search_list = ["N", "SLT", "MM"]result_list = [code_to_text[code] for code in search_list]
Both of these give the same value of result_list:
['NONE', 'SLIGHT', 'MEDIUM']
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